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2 years ago

8

A diver runs horizontally with a speed of 1.20 m/s off a platform that is 10.0 m above the water. What is his speed just before

striking the water

1 answer:

DENIUS [597]2 years ago

4 0

Answer:

14.05m/s

Explanation:

We are given that

Horizontal speed of diver= $v_x=1.2m/s$

Distance, y=-10 m

It is taken as negative because the diver goes downward.

There is no air resistance therefore, there is acceleration due to gravity .

We know that

$g=-9.8 m/s^2$

Initial vertical velocity, $u_y=0$

$v^2-u^2=2as$

Using the formula

$v^2_y-0=2\times (-9.8)(-10)$

$v_y=\sqrt{2(-9.8)(-10)}$

$v_y=14 m/s$

$v=\sqrt{v^2_x+v^2_y}$

Using the formula

$v=\sqrt{(1.2)^2+(14)^2}=14.05m/s$

Hence, the speed of diver before just striking the water=14.05m/s

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Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave r

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We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon

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Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

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2 years ago

Read 2 more answers

1. What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?

Anit [1.1K]

Answer:

<h3>The answer is 4.53 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

<h3>4.53 kgm/s</h3>

Hope this helps you

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2 years ago

A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angu

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Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let $\omega$ is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{\omega-0}{t}$

$\alpha =\dfrac{\omega}{t}$

$t=\dfrac{\omega}{\alpha }$ ............(1)

Using second equation of kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$\theta=\dfrac{1}{2}\alpha t^2$

Using equation (1) in above equation

$\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

In one revolution, $\theta=4\pi$ (in 2 revolutions)

$4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

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2 years ago

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Dee is on a swing in the playground. the chains are 2.5 m long, and the tension in each chain is 450 n when dee is 55 cm above t

Leno4ka [110]

Refer to the diagram shown below.

From the geometry, obtain
x = 2.5 - 0.55 = 1.95 m
cos θ = 1.95/2.5 = 0.78
θ = cos⁻¹ 0.78 = 38.74°

From the free body diagram, the tension in the chain is 450 N.
F is the centripetal force,
W is Dee's weight.

The components of the tension are
Horizontal component = 450 sin(38.74°) = 281.6 N, acting left.
Vertical component = 450 cos(38.74°) = 351.0 N, acting upward.

Answers:
Horizontal: 281.6, acting left.
Vertical: 351.0 N, acting upward.

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2 years ago

A metal sphere with radius R1 has a charge Q1. Take the electric potential to be zero at an infinite distance from the sphere.Ex

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Answer:

E = k*Q₁/R₁² V/m

V = k*Q₁/R₁ Volt

Explanation:

Given:

- Charge distributed on the sphere is Q₁

- The radius of sphere is R₁

- The electric potential at infinity is 0

Find:

What is the electric field at the surface of the sphere?E.

What is the electric potential at the surface of the sphere?V

Solution:

- The 3 dimensional space around a charge(source) in which its effects is felt is known in the electric field.

- The strength at any point inside the electric field is defined by the force experienced by a unit positive charge placed at that point.

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F = k*Q₁/R₁²

- Then the electric field at that point is

E = F/1

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V = k*Q₁/R₁ Volt

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